Verified answer

Ответ:

[tex]y=\sqrt[3]{x^2+tgx+15}=\Big(x^2+tgx+15\Big)^{\frac{1}{3}}[/tex]

Формула:  [tex](u^{n})'=n\cdot u^{n-1}\cdot u'[/tex]  .

[tex]y'=\dfrac{1}{3}\cdot \Big(x^2+tgx+15\Big)^{-\frac{2}{3}}\cdot \Big(2x+\dfrac{1}{cos^2x}\Big)=\dfrac{1}{3\sqrt[3]{(x^2+tgx+15)^2}}\cdot \Big(2x+\dfrac{1}{cos^2x}\Big)[/tex]